Guest Post by Willis Eschenbach
Short Post. You can skip this if you understand the tidal force. Some folks seem confused about the nature of tidal forces. Today I saw this gem: “The tide raising force acts in both directions (bulge on each side in the simplistic model)” … the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.
To clarify the situation, let me give an example. Imagine three one-kilogram masses M1, M2, and M3, connected by a blue rope, that are in free-fall directly into the sun. Figure 1 shows the situation.
Figure 1. Forces on three one-kilogram masses M1—M3, which are in free-fall straight towards the sun, and are connected by the blue rope. The equations at the bottom show the force of gravity on each of the three masses. The “tidal force” is the difference between two gravitational forces on two objects. Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons), and the lunar tidal forces are about twice that, about a micronewton. By comparison, Earth’s gravity exerts a force of about 10 newtons on a 1-kg mass at the surface …
Clearly, the gravitational force on M1 is greater than the force on M2, which in turn is greater than the force on M3. The difference between those two gravitational forces is the tidal force. There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. As a result of these tidal forces, as the three masses fall towards the sun. the near mass M1 moves the furthest, the far mass M3 moves the least, and the blue rope is always under tension.
For tidal forces on a theoretical planet, just imagine that the surface of the planet is the gray dotted line, and that “r” is the radius of the planet. The tidal force is the difference between the force of gravity on the unit mass M2 at the center of the planet, and the forces of gravity on the unit masses M1 and M3 at the points nearest and furthest from the sun.
Calculating the Tidal Forces
As shown in Figure 1, to calculate the tidal force you can just subtract the gravitational force on the unit mass at the center of the planet from the force on the other unit mass, for example GF1 – GF2. So we can use the equations in Figure 1 to calculate the force.
However, when the distance D is much, much larger than the radius r, we can make some simplifying assumptions. First, we can assume that the force between M1 and M2 is the same as the force between M2 and M3. It’s not, as you can see from the equations … but it is not a large error. For tides from the moon, the individual tidal forces are about ±5% different from their average value.
The next simplifying assumption we can make if D is much larger than r is that the average tidal force can be calculated as
Tidal Force = 2 * G * sunmass * r / D^3
This leads to tiny errors with respect to the sun’s average tidal force on the earth, and slightly larger errors with respect to the moon. For the sun we have
Near side true tidal force on earth (GF1 – GF2) = 0.50590 µN
Far side true force (GF2 – GF3) = 0.50583
True average = 0.50587
Approximation = 0.50587
So the approximation of the average tidal force is good to five significant digits …
And for the moon we have:
Near side true tidal force on earth (GF1 – GF2) = 1.1350 µN
Far side true force (GF2 – GF3) = 1.0796
True average = 1.1073
Approximation = 1.1067
For the moon, the approximation of the average is good to four significant digits. For most results, of course, those differences are far too small to be meaningful, and we can use the approximation without concern.
Why Are There Two Tidal Bulges?
Moving on, let’s look at why there are tidal bulges on both the near and far sides. It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun. So lets look at Figure 1 again, and this time we’ll include a planet and an ocean. As in Figure 1, the planet is simply free-falling into the sun. That result is shown in Figure 2:
Figure 2. Tidal forces elongating the planet and the ocean. Note that the planet is elongated as well, but this is not shown in the diagram because obviously, tides in the solid planet are much smaller than tides in the ocean. NOTE THAT THIS PLANET IS NOT THE EARTH.
As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet. And the planet, in turn is pulled more than the far side, so it pulls away from the ocean. These relative motions create the “bulges”
In other words, the “bulges” on the two sides of the planet are simply a result of the tidal forces stretching the entire system. Indeed, the atmosphere is subject to the same phenomenon, so there are atmospheric tides as well. In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner. However, such connections, while certainly possible, have proven to be elusive and difficult to establish. One complicating factor is that the atmospheric tides that are caused by the sun occur in combinations with solar heating, which makes the atmosphere swell. Also, the variations in pressure caused by the atmospheric tides are small, on the order of a tenth of a millibar (or of a hectopascal), or a variation of only 0.01% …
Finally, I’ve said that the situation in Figure 2 shows that planet free-falling into the sun. As it gets closer and closer, the tidal forces stretching the system get greater and greater, stretching the system more and more. Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces … dang, to have a ring-side seat right on the Roche limit line for that spectacle would be awesome, even if it were perhaps a bit warm …
My best to all,
w.
PS—As with gravity, every mass in the universe exerts a tidal force on every other mass, so you can personally take credit for raising a tide on the surface of the sun … yeah, it’s a small tide, but it’s real. Just sayin’, I had someone claim the opposite. Lots of tidal misinformation out there.
THE USUAL: If you disagree with me or anyone, please quote exactly what it is you disagree with, and tell us precisely why you disagree. That way, we can understand exactly what it is you object to.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
A lot of the misconceptions on this thread arise from people trying to apply Newton’s laws in an accelerated coordinate system. Newton’s laws hold only in an inertial coordinate system. In an accelerated coordinate system, you must take account of the acceleration with respect to inertial space.
The Earth is accelerating towards the Sun. So, to get the net specific force acting at the surface of the Earth, you must subtract out the Earth’s acceleration towards the Sun. As I showed how to do up above, where I showed that the true effect of the centripetal acceleration is to increase the specific force acting on the near and the far side by 50%. This is for solar induced tides only, as only there is the distance to the barycenter approximately equal to the distance to the Sun. The effect is much less for lunar induced tides.
Bart says:
February 16, 2014 at 12:27 pm
I didn’t answer because it is already answered. The tidal effect for falling straight in is missing the additional component due to centripetal acceleration, so its a factor of 2 instead of 3.
Stop being stupid. Read the equations I provided. They are correct.
The issue being of concern was that of falling straight in and centripetal acceleration for that is not a concern, so you can keep your snide comments up where they belong. The correct derivation for the general case is the one I presented in slide 21 of http://www.leif.org/research/AGU%20Fall%202011%20SH34B-08.pdf
lsvalgaard says:
February 16, 2014 at 1:01 pm
You are fixated on your interpretation of tides on the Sun. I do not have a problem with that interpretation. As I have stated time and time again, centripetal effects are minscule there, because the barycenter is quite close to the center of the Sun.
The place where centripetal effects are not insignificant is solar induced tides on the Earth. I keep explaining this, and it seems you simply disregard it, and continue on, as though I am claiming something I am not. If “stop being stupid” is snide, then let me clarify: stop telling me I am wrong about something I am not claiming.
Bart says:
February 16, 2014 at 12:39 pm
“I was stating my position, not imputing yours. We are in effing agreement on this. Why are you arguing?”
Because you then say
““centripetal/centrifugal forces are almost totally irrelevant in tidal calculations.”
No, they are not. There are not the largest contributor, in fact are insignificant for any but solar induced tides on the Earth. And, they are NOT the cause of the symmetry in the bulge.”
So my “totally irrelevant ” translates into “No, they are not. There[sic] are not the largest contributor”
Who is disagreeing with whom?
—-
From the link I provided
Tidal trivia.
•Amplitude of gravitational tides in deep mid-ocean: about 1 meter.
•Shoreline tides can be more than 10 times as large as in mid-ocean.
•Amplitude of tides in the Earth’s crust: about 20 cm.
•The gravitational force of sun on Earth is 178 times as large as the force of moon on Earth.
•Ratio of sun/moon tidal forces on Earth is 0.465.
•Tidal stretch of human body changes its height by fraction 10-16, an amount 1000 times smaller than the diameter of an atom. By comparison, the stress produced by the body’s own weight causes a fractional change in body height of 10-2. [Sawicki]
•Tidal friction causes Earth days to lengthen 1.6 milliseconds/century. [Sawicki]
•Angular velocity of Earth’s axial rotation: 7.29 x 10-5 rad/s.
•Angular velocity of moon’s revolution around Earth: 2.67 x 10-6 rad/s.
•Earth polar diameter: 12710 km.
•Earth equatorial diameter: 12756 km.
•Difference between these diameters: 46 km.
•Difference between these radii: 23 km, or 0.4 %.
•Thickness of atmosphere, about 100 km.
•Centripetal acceleration at Earth’s surface due to Earth’s axial rotation: 0.034 m/s2
•Size of centripetal acceleration at Earth’s surface due to Earth’s circular motion around the Earth-moon barycenter: 3.3 x 10-5 m/s2.
RichardLH says:
February 16, 2014 at 1:12 pm
“•Centripetal acceleration at Earth’s surface due to Earth’s axial rotation: 0.034 m/s2”
Not talking about axial rotation. I am talking about revolution about the Sun.
“•Size of centripetal acceleration at Earth’s surface due to Earth’s circular motion around the Earth-moon barycenter: 3.3 x 10-5 m/s2.”
Earth-Moon>/b> barycenter. To the degree that the Earth’s orbit about the Sun is a circle, the centrifugal “force” precisely cancels the gravitational force. That is why you experience free-fall in orbit – there are zero net forces. For solar induced tides, the sensitivity to position away from the center of gravity of the Earth from centrifugal “force” is 1/2 of the sensitivity from gravitational force (I showed this up above). That is why the centrifugal “force” adds 50% to the gravitational forces affecting the tidal bulge.
The sensitivity from centrifugal “force” for lunar induced Earthly tides is much smaller, because the barycenter is actually within the Earth itself, a small fraction of the distance to the Moon.
I believe that most of the debate in this thread is due to the confusion between reference frames. Let’s consider the simple case of three small asteroids falling directly towards the sun but separated along a radial distance, following Willis’ model. Assume that the asteroids are small enough so that we can ignore gravitational attraction between them. Our point of view is on an observational platform a fixed distance from the sun, held there by rockets let us say, and our weight represents the gravitational attraction of the sun. We see the three asteroids falling towards the sun and separating as they fall due to the gradient in the gravitational field. Each has a different acceleration at any point in time due to the different radial distance of each. There is no tidal force; there is just three objects falling. Now add a cable connecting the two outer asteroids to the centre one and maintaining a fixed distance between them. The cables will be under tension so as to counteract the gravitational gradient of the sun. So there are only two forces, the sun’s gravity and the tension in the cables. There is nothing that we can call a tidal force.
Now let’s consider a different frame of reference and place ourselves on the centre asteroid. Here we are in free fall towards the sun and are thus weightless and so assume that there is no gravitational force, but we must explain why the two outer asteroids are moving away from us in opposite directions (back to the no cable case). We must invent a fictitious force to explain this. One that draws objects away from us in the two opposite directions and which we call the tidal force. Put the cable back and this force is balanced by the tension in the cable. It is only in this frame that we can properly talk of tidal forces.
With the revolving and rotating earth the situation is more complicated and there are other fictitious forces (centrifugal, Coriolis, etc) which make the calculations more complicated depending on the frame that we choose to view things. This is shown in detail in the Sirtoli link as given by yirgach. We also have the Earth’s own gravitational field, the moon as well as the sun, and the tensile strength’s of land and sea.
So we either have a gradient in the gravitational field(s) or fictitious tidal forces, but not both. I like this comment of Sirtoli “Those forces are called fictitious because they have no physical reality, they emerge only in accelerating reference frames. The only forces worthy of being called “real” are the four fundamental forces: the gravitational, electrical, strong nuclear and weak nuclear. No physical effect can derive from a fictitious force: we can always choose a reference system in which this force will disappear.” I wouldn’t put it quite this starkly but I believe that it supports my position that the so called tidal force is fictitious.
The bottom line: I think that Willis is using the term tidal force somewhat incorrectly although his explanation of the tidal effect is correct.
Let me clarify my position one last time, before I must depart.
When I say that the tidal bulge is symmetric due to either gravity alone, or centripetal acceleration alone, or both together, I mean that it is two-sided – it bulges on the near side AND on the far side, and both contributions result in symmetric, two-sided bulges.
The centripetal acceleration is not necessary for the bulge to be two-sided. It only amplifies the two-sided bulge beyond what it would otherwise be. It amplifies the specific force differential on the near and far side by near 50% when the distance to the barycenter is close to the distance between the bodies.
As the relationship between that force differential and the size of the bulge is nonlinear, that 50% increase does not result in a 50% increase in maximum bulge. But, the effect is not insignificant.
Bart:
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
A couple of orders of magnitude difference as I suggested before
Therefore
Centripetal acceleration at Earth’s surface due to Earth’s axial rotation = 0.034 m/s2
Centripetal acceleration at Earth’s surface due to Earth’s orbit around the Sun ~ 0.034 * (1.99/7.29)^-2 m/s2
or did I get something wrong?
EDIT: Damn it – fingers again
~ 0.034 * (1.99/7.29) * 10^-2 m/s2
Mike M says:
February 16, 2014 at 6:00 am
Oh you’re welcome,
I think you asked a very clear and logical question and you deserve an answer to it.
However, as I see it, you are wrong because the rotational effect from the Earth is cancelled by other effects.
You look at the speed on the Earth’s surface and add that to the orbital velocity at midnight and subtract it at noon. Very logical and intuitive, but I think it is wrong, because if you choose to look at the orbital effect that way you also have to take into account the orbits effects on the centripetal acceleration on the Earth’s surface.
An object on the Earth’s surface will have smaller angular velocity at noon than at midnight. At noon the arc of the Earth’s surface is in the opposite direction as the orbital arc. The angular velocity from the orbit is therefore cancelling out some of the angular velocity from the Earth’s rotation. This causes a reduction in the centripetal acceleration from the Earth’s rotation at noon.
At midnight they are in the same direction and you have the opposite effect.
This effect cancels out the change in orbital velocity.
Concerning your discussion on varying gravitation during the day, I think you are partially right; the Earth’s gravity is not absolutely constant around the clock. It varies with the tides but the variation is minuscule.
/ Jan
lsvalgaard says:
February 15, 2014 at 9:12 am
According to your ‘theory’ a pendulum clock should vary its ticking rate during the day and it does not.
==============
Why is that? We know that pendulum clocks loose time if moved to a higher elevation. According to wikipedia 16 seconds/day at 4000 meters altitude.
It would seem from this result that a pendulum clock on the equator would locally tick slowest during a full eclipse of the sun. Thus the pendulum clock should gain or lose time, depending on the tidal forces from the sun and moon. Though the effect might be small.
Since gravity and acceleration are equivalent, the same argument can be applied to the rotation of the earth. Pendulum clocks should run slower at the equator than the poles, though this is complicated by the equatorial bulge of the earth.
This paper calculates that a pendulum clock varies about 8ms/day due to the tides.
http://leapsecond.com/hsn2006/pendulum-tides-ch1.pdf
http://leapsecond.com/hsn2006/pendulum-tides-ch2.pdf
And in this paper the previous author examines the Allan variance using data from Pierre Boucheron’s mid-1980’s timing experiment with a Shortt free pendulum clock.
“Conclusion
With the best of conditions the effect of tides on
pendulum clocks can be seen directly in rate or
progressive time error charts (e.g., Fedchenko gravim
eters via George Feinstein). We have also
seen the effect of tides through F
ourier spectral analysis (e.g., Philip Woodward). In this section
we explored how an Allan deviation plot can be
used to see the effect of tides on a precision
pendulum clock.”
http://en.wikipedia.org/wiki/Shortt-Synchronome_clock
Recent accuracy measurement
In 1984 Pierre Boucheron studied the accuracy of a Shortt clock preserved at the US Naval Observatory.[3][17] Using modern optical sensors which detected the precise time of passage of the pendulum without disturbing it, he compared its rate to an atomic clock for a month. He found that it was stable to 200 microseconds per day, equivalent to an error rate of one second in 12 years, far more accurate than the 1 second per year that was previously measured. His data revealed the clock was so sensitive it was detecting the slight changes in gravity due to tidal distortions in the solid Earth caused by the gravity of the Sun and Moon.[18]
Canute Ponders The Tides Posted on February 14, 2014 by Willis Eschenbach
Short Post. You can skip this if you understand the tidal force.
A better heading would be Canute ponders the tidal force as the tides as commonly understood on earth alone are infinitely complicated.
Sticking to your analogy of the 3 1 Kgm masses in free fall into the sun.
presumably all launched at the same time , at the same initial velocity
You are basically saying the difference in distance between them will increase over time as they get closer to the sun as the gravity differential is different proportional to the distance between them.
If this is the radius of the earth for a sol sized attractor and they are starting at the distance of the earth from the sun.
The speed of the fall would have to be equivalent to whatever speed is keeping the earth in orbit so it would be pretty fast.
I don’t know if there would be time for a tide differential to develop before the earth would crash into the sun.
So the tide differential shown at the time we are discussing would have to match some previous condition to get us to this point in time.
At which previous time the 3 balls would have to have been released at different speeds
A bit like the tortoise and the hare really, I need better maths with these infinite regress problems that I don’t have yet.
So, its not a short post, Its not simple.
Centrifugal, centripetal forces are irrelevant for this post only.
I now understand that when ball B gets to where ball A was, Ball A is slightly further apart from ball B by Richard’s delta amount due to the fact that it has had a slightly stronger G working on it over that distance and has therefore traveled further.
This is called a tidal force in this thread but would be basically the force of different strength gravity.
I f the gravity was the same for all 3 balls they would never move apart no matter how fast they all were accelerating.
I don’t understand how one can be falling but not moving [as in orbit] thus giving a tidal force.
It like saying yes the gravity difference is working but there is no gravity angech 117/2/2014.
It might need the explanation of centripetal forces which is a way of saying infinitely falling and accelerating but not moving in position.
angech says:
February 16, 2014 at 3:38 pm
I don’t know if there would be time for a tide differential to develop before the earth would crash into the sun.
=========
light takes 8 minutes to make the journey. free fall time from earth to sun is 65 days.
angech says:
February 16, 2014 at 3:38 pm
I don’t understand how one can be falling but not moving [as in orbit] thus giving a tidal force.
==========
the earth is moving while in orbit. in effect, the earth is always “falling” towards the sun, but at the same time moving sideways fast enough so that it never hits the sun.
for example, face an object and take 2.5 steps backwards. This is your starting position. Now, take 1 step towards the object, then two steps sideways. If you originally started 2.5 steps from the object, you will end up still 2.5 steps from the object. Turn slightly to face the object. Take 1 step towards the object and 2.5 steps sideways, you are still 2.5 steps from the object. Repeat over and over, you are in orbit.
Jan Kjetil Andersen says: February 16, 2014 at 2:13 pm “.. I think it is wrong, because if you choose to look at the orbital effect that way you also have to take into account the orbits effects on the centripetal acceleration on the Earth’s surface.”
A centripetal force is one which acts solely perpendicular to the direction of a particle’s velocity at any given instant. Such can be gravity OR something mechanically pushing on it, (such as a road surface pushing sideways on a tire making the car turn). So, I’m unclear exactly what you are describing with – “the orbits effects on the centripetal acceleration on the Earth’s surface”?
“The angular velocity from the orbit is therefore cancelling out some of the angular velocity from the Earth’s rotation.” Easily small enough to ignore for the purpose of illustration. (And I think it adds to it. If earth was like mercury it would complete one revolution per year. Now start rotating it to get one full day per year and you get 2 revolutions – not zero.)
“This effect cancels out the change in orbital velocity.” For all of earth it does but no, we’re talking about the affect it has on individual particles such as molecules of water that will experience a net accelerational difference oscillating throughout the day and therefore react to that change as opposed to if the earth rotated once per year so all the particles experience a balanced solar gravity = V^2/R at all times, (not accounting for Willis’ affect).
“Pendulum clocks should run slower at the equator than the poles, though this is complicated by the equatorial bulge of the earth.” That’s true but that’s a uniform centrifugal effect throughout the day and therefore, as a constant, makes no difference to the effect I’m addressing. Going slower at noon and faster at midnight WRT to solar orbital velocity, at the equator, will cause the pendulum to go slowest at noon and at midnight by a maximum amount on the equator also decreasing to zero at either pole for the same reason that your the centrifugal effect does, because: V= Ω*(Requator) * cos(latitude). Both effects rely on earth’s rotation but one has nothing to do with the other.
William Sears says: February 16, 2014 at 1:35 pm
“Here we are in free fall towards the sun and are thus weightless and so assume that there is no gravitational force, but we must explain why the two outer asteroids are moving away from us in opposite directions (back to the no cable case). We must invent a fictitious force to explain this.”
False. Acceleration due to gravity does not mysticaly ‘go away’ just because you are in free fall – it acts upon the object no matter which way it goes or (excepting relativistic velocities) how fast.
So there is no fictitious force per your example – all three are being accelerated by a REAL gravitation force that varys per asteroid by how close it is to the sun.
Mike M, you say that because you are still viewing the situation in the first reference frame of my example. You have to put yourself on the falling asteroid. Your objection can be used to dismiss any fictitious force by simply refusing to put yourself in the non-inertial frame.
William Sears says: February 16, 2014 at 5:37 pm “You have to put yourself on the falling asteroid. ”
Okay – I’m on your asteroid & the sun is accelerating at me. So what? The sun’s gravity is still acting upon my asteroid and it upon the sun. THAT doesn’t go away in ANY interial frame of reference.
Everything you wanted to know about tides, but were afraid to ask …
http://www.lhup.edu/~dsimanek/scenario/tides.htm
via The Chiefio ..
“Mike M says:
February 15, 2014 at 12:15 pm
Let’s not forget that earth’s crust is extremely thin WRT earth’s diameter and is going to flex to “bulging forces” mostly at the equator along with the ocean. Could that explain why ocean tide varies the least at the equator and greatest at the higher latitudes by virtue that those regions of crust are not being bulged as much making the determination of which way is “downhill” at various times of the day rather complex.. ??”
Given the 299 comments to this blog I am not going to try to find if anyone has actually replied to your post – I could not find a response over the next 50 or so. However, that the ocean tides vary lest (in very general terms) at the equator is not due to the earth’s crust also bulging. The tides are created by the sun and moon, and to a minimal extent by other planets. But the displacement on the earth’s surface is greatly affected by the shape of the oceans and seas, both shore lines, continental undersea extensions and the ocean bottoms. A very important point is the natural resonance of the oceans. This could be any reasonable period but logic suggests that periods close to the moon’s or sun’s maximum/minimum attraction periods will tend to result in greatest effects, as will periods a multiple or submultiple of those timings. Thus if the natural period of a body of water is 12 ( or 6) hours it will resonate with the sun and solar tides will be important. If the natural period is 12.5 or 6.25 hours moon tides will be important.
Then think of what happens when you are holding a bowl full of water. As you struggle to hold it up you move the bowl slightly, and the water slops about. At the resonant frequency of the bowl + water it will slop about a lot, at a frequency of your shaking substantially different the slopping may be minimal. But notice what happens in the middle of the bowl. Water moves from side to side but not up and down. At the edges water moves up and down but not from side to side. If your hands are at the sides, water vertical movement will also be greater close to your hands, and minimal well away from your hands. Sea and ocean basins act like this.
Tide ranges are high in places like the Bay of Fundy, the Thames, the Hoogly because they are at the edge of the seas/oceans, and because the shape of the land magnifies the vertical movement. At mid-Pacific islands tidal ranges are low – water tends to just move very slightly sideways rather than up and down. The Mediterranean has a very low tidal range as it is just not the right shape for resonance, and you have Italy, Sicily, Malta, and the Greek islands all acting to damp any horizontal water movement. Port Philip is an interesting case – a near landlocked body of water with a single narrow entrance. From half tide through to high tide and on to half tide outside the entrance, water pours in through The Rip, and the tide rises inside Port Philip, at Port Melbourne, Geelong, etc. For the other half of the tidal cycle outside, through low water, water pours out, and the tide inside Port Philip falls. So high water inside and outside are out of phase by six hours, roughly.
“Willis Eschenbach says:
February 15, 2014 at 11:30 pm
This has a maximum at 45°N and 45°S, which is one reason why the tides are big in those latitudes. However, the maximum is only half of the tidal force [ sin(45) * cos(45) = 0.5 ].”
OK so the tidal forces are zero at the poles. Unfortunately your equation also means the force is zero at the equator. Remember Sin(0) * cos (0) = 0.
Perhaps this needs to be modified or explained better?
Mike M, “so what?” isn’t much of an argument and therefore I don’t see how I can help you, but I will give it one more try. Remember that in any reference frame you can only use what you can measure locally and that you can not look outside in an attempt to circumvent this, no matter how tempting. What the sun appears to be doing is irrelevant. Block it with a dust cloud if it bothers you. You are weightless and there is no local experiment that you can do to detect the existence of a gravitational field. But the other two asteroids are accelerating away from you. Explain this without bringing in knowledge that you do not have in this reference frame. Forget about any curvature of space arguments as we are dealing with classical physics here.
Mike M says:
February 16, 2014 at 4:39 pm
Going slower at noon and faster at midnight WRT to solar orbital velocity
============
the orbital speed of the earth around the sun is simply a change of reference field. Since the centrifugal force is exactly opposite the centripetal force (gravitational attraction of the sun) you can get the correct answer using the centrifugal force in place of the gravitational force.
As a tidal comparison consider two soft rubber balls that are mounted through their centres to rigid posts close to each other. Opposite electric charges are deposited on them so that there is a strong electrostatic attraction. There is no charge on the posts and assume that the charge on the rubber balls is uniformly distributed and immobile. The shape of the rubber balls will be distorted due to this attraction and the fact that they are pinned through their centres. But the distortion will only be towards the other rubber ball, stretched out on the near side and compressed on the far side. A nonlinear field is not even necessary in this case. A similar effect is seen by pinning a plasticine ball to a wall and letting gravity do the stretching.
I think that this is the subconscious picture in most people’s minds in trying to understand the tides and why the stretching occurs in two directions. This may also lead to the perceived need to use a centrifugal force explanation. It also shows why the celestial object must be in free fall (eg orbit or the direct falling example used by Willis) to get the tidal stretch in two directions.